Some inequality I stumbled upon, $b^3-a^2b+2a \geq 0$

Question

$b^3-a^2b+2a \geq 0$

Given: $(a,b) \in \mathbb{N}$ and a>b. Find when this holds. This is an inequality I stumbled upon while I was surfing, and like any other day, I sat down trying to think about it. In general, playing with it gives me that $b \geq 3$ "guarantees" no value of $a$. Although proving the statement formally seems to take me forever. Initially, I tried to bring it to quadratic form in $a$, and if $b$ behaves like a constant, then we can see $P(a) \geq 0$ which imposes $D \leq 0$, but that implies $b^4+1 \leq 0$ which is absurd in the field we are working on. This somehow convinced me to try working on the cubic which doesn't seem to be factorizable either.

I would love a solution that could include the remedy and why my approach didn't work.

Thank you.

Answer 1

First, for the error in your reasoning:

Note that (with $b$ fixed) solving for $P(a) \geq 0 $ does not impose $ D \leq 0$.
That is true only if $P(a) \geq 0 $ for all values of $a$ (which the problem doesn't require).
In this case, substantiated by your observation that $ D = 1+b^4 > 0 $, we know that for each $b$, there will be some positive and negative values of $P(a)$.
To find which are the valid $a$ values, we have to find the roots of the quadratic equation. This is essentially shown by Prasiortle.

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Now, for a quick solution:

Write the inequality as $ \frac{2}{b} \geq \frac{ a^2 - 1}{a}$.
Observe that for $ a \geq 2$, we always have $ a < a^2 - 1$.
Hence, for this inequality to hold true, we must have $ \frac{2}{b} \geq \frac{ a^2 -1 }{ a} > 1 $, or that $ b =1$.
Thus, this gives us $ 2a \geq (a^2 -1 )$, or that $ a = 2$.

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Now, for my original solution using a different method.

Let $ a = b + k$ where $ k$ is a positive integer.

Then the inequality becomes

$$ 2b + 2k \geq 2b^2k + bk^2. $$

From here, since

• $2b < bk^2$ except when $ k = 1$
• $2k \leq 2b^2k$

It follows that we must have $k = 1$. This then gives us

$$2b + 2 \geq 2b^2 + b $$

and solving the quadratic shows us that the only positive integer solution is $ b = 1$.

Hence, the only solution is $ (a, b) = (2, 1)$.

Answer 2

With $a, b \in \mathbb{N}$ and $a > b$, the only solution is $a = 2, b = 1$. To prove this, we rewrite the inequality as $$ba^2-2a-b^3 \leq 0,$$ with $a^2$ coefficient $b$, which is positive, and discriminant $4\left(b^4+1\right)$, which is also positive. Thus the solution over $\mathbb{R}$ is $$\frac{2-2\sqrt{b^4+1}}{2b} \leq a \leq \frac{2+2\sqrt{b^4+1}}{2b},$$ i.e. $$\frac{1-\sqrt{b^4+1}}{b} \leq a \leq \frac{1+\sqrt{b^4+1}}{b}.$$ Imposing the constraints $a > b$ and $a \in \mathbb{N}$, we require there to be an integer strictly greater than $b$ and less than or equal to $\frac{1+\sqrt{b^4+1}}{b}$. Since $b$ itself is a positive integer, this means the latter must be at least $b+1$, so $$\begin{align*}\frac{1+\sqrt{b^4+1}}{b} \geq b+1 &\iff 1+\sqrt{b^4+1} \geq b^2+b \qquad \text{(as } b > 0 \text{)} \\ &\iff \sqrt{b^4+1} \geq b^2+b-1.\end{align*}$$ If $b^2+b-1 \leq 0$, then this will automatically be true; but it's easily checked that there is no such $b \in \mathbb{N}$. Therefore, both sides will necessarily be positive, so we can square both sides to give $$\begin{align*}&b^4+1 \geq b^4+2b^3-b^2-2b+1 \\ &\iff 2b^3-b^2-2b \leq 0 \\ &\iff 2b^2-b-2 \leq 0 \qquad \text{(as } b > 0 \text{)} \\ &\iff \frac{1-\sqrt{17}}{4} \leq b \leq \frac{1+\sqrt{17}}{4} \\ &\iff b = 1 \qquad \text{(as } b \in \mathbb{N} \text{)}.\end{align*}$$ Finally, setting $b = 1$ in the original inequality yields $$\begin{align*}a^2-2a-1 \leq 0 &\iff 1-\sqrt{2} \leq a \leq 1+\sqrt{2} \\&\iff a \in \{1,2\} \qquad \text{(as } a \in \mathbb{N} \text{)}\end{align*}$$ but since $a > b$, only $a = 2$ is a valid solution, completing the proof.